Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{-2k - 8}{-k + 5} \div \dfrac{3k^2 + 18k + 24}{2k^2 + 12k + 16} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{-2k - 8}{-k + 5} \times \dfrac{2k^2 + 12k + 16}{3k^2 + 18k + 24} $ First factor out any common factors. $z = \dfrac{-2(k + 4)}{-(k - 5)} \times \dfrac{2(k^2 + 6k + 8)}{3(k^2 + 6k + 8)} $ Then factor the quadratic expressions. $z = \dfrac {-2(k + 4)} {-(k - 5)} \times \dfrac {2(k + 2)(k + 4)} {3(k + 2)(k + 4)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-2(k + 4) \times 2(k + 2)(k + 4) } {-(k - 5) \times 3(k + 2)(k + 4) } $ $z = \dfrac {-4(k + 2)(k + 4)(k + 4)} {-3(k + 2)(k + 4)(k - 5)} $ Notice that $(k + 2)$ and $(k + 4)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-4\cancel{(k + 2)}(k + 4)(k + 4)} {-3\cancel{(k + 2)}(k + 4)(k - 5)} $ We are dividing by $k + 2$ , so $k + 2 \neq 0$ Therefore, $k \neq -2$ $z = \dfrac {-4\cancel{(k + 2)}(k + 4)\cancel{(k + 4)}} {-3\cancel{(k + 2)}\cancel{(k + 4)}(k - 5)} $ We are dividing by $k + 4$ , so $k + 4 \neq 0$ Therefore, $k \neq -4$ $z = \dfrac {-4(k + 4)} {-3(k - 5)} $ $ z = \dfrac{4(k + 4)}{3(k - 5)}; k \neq -2; k \neq -4 $